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Is it to find the zeros in binary format?
If yes then:

count = 0
while (n>0) {
count = count + !(n&1)
n=n>>1 //Right shift by 1
}
return count

To find in decimal format the

count = 0
while (n>0) {
count = count + (n%10>0 ? 0 : 1)
n = n/10;
}
return count

while(num!=0)
    {
        k=num%10;

        if(k==0)
        count++;


        num=num/10;

    }
    cout<<"\n"<<count;

It can be found by finding number of times 5 appears in the factors, since 5 *2 always gives one zero. So to proceed, try to find maximum value of n in 5^n in case where number is divisible.
Ex: 100 => 5^2 , therefore n = 2 so maximum trailing zeros are 2.

int findZeroBit(int num)
{
int n=1,count=0;
while(n <= num)
{
if(!(n & num))
count++;
n<<=1;

}
return count;
}

int findZeroBit(int num)
{
int n=1,count=0;
while(n <= num)
{
if(!(n & num))
count++;
n<<=1;

}
return count;
}

I would it by counting the number of ones and subtracting from sizeof(integer).

Assume input = v

for (int c =0; v; v>>1)
c+= (v&1)

At end 'c' will contain number of ones in number

So number of zeros is:

NoOfZeros = sizeof(int) *8 - c

(8= number of bits in byte. I guess this is an assumption so the code may not be fully portable)

So final code

int NoOfZeros(int v)
{
for (int c = 0; v; v>>1)
c+= (v&1);

return sizeof(int)*8-c;
}

int FindZeros(uInt val)
{
static uInt cnt = 0, loop = 0;
if(loop == 32)
{
return cnt;
}
loop++;
if(!(val & 1))
cnt += 1;
FindZeros(val >> 1);
}

int count_zero_bits(int n)
{
   int x = ~o;
   x=x^n;    // this will mask the bits that are set to 1, 
   int count=0;
  while(x!=0)
   {
      if(x&1==1)
       {
           count++;

       }
   x=x>>1;
  
   }

 return count;
}

int count_zero_bits(int n)
{
   int x = ~o;
   x=x^n;    // this will mask the bits that are set to 1, 
   int count=0;
  while(x!=0)
   {
      if(x&1==1)
       {
           count++;

       }
   x=x>>1;
  
   }

 return count;
}

n = n&(n-1) : set last right most 1 to 0;
count number of 1's and subtract from (sizeof(int)*8)

int numberofzeros(int n)
{
int count =0;
while(n)
{ count++;
n= n&(n-1);
}
return ((sizeof(int)*8)-count);

}

#include <iostream>
using namespace std;
int findTrailingZeros(int n)
{
int count = 0,sum=0,res=0;

while(n>5)
{
res=n/5;
sum=sum+res;
n=res;
}
return sum;
}
int main()
{
int n,res;
cin>>n;
res=findTrailingZeros(n);
cout<<res;
return 0;
}

#include <iostream>
using namespace std;
int findTrailingZeros(int n)
{
int count = 0,sum=0,res=0;

while(n>5)
{
res=n/5;
sum=sum+res;
n=res;
}
return sum;
}
int main()
{
int n,res;
cin>>n;
res=findTrailingZeros(n);
cout<<res;
return 0;
}

int findTrailingZeros(int n)
{
int count = 0,sum=0,res=0;

while(n>5)
{
res=n/5;
sum=sum+res;
n=res;
}
return sum;
}
int main()
{
int n,res;
cin>>n;
res=findTrailingZeros(n);
cout<<res;
return 0;
}

int findTrailingZeros(int n)
{
int count = 0,sum=0,res=0;

while(n>5)
{
res=n/5;
sum=sum+res;
n=res;
}
return sum;
}
int main()
{
int n,res;
cin>>n;
res=findTrailingZeros(n);
cout<<res;
return 0;
}

#include <iostream>
using namespace std;
int findTrailingZeros(int n)
{

int count = 0,sum=0,res=0;

while(n>5)
{
res=n/5;
sum=sum+res;
n=res;
}
return sum;
}
int main()
{
int n,res;
cin>>n;
res=findTrailingZeros(n);
cout<<res;
return 0;

}

#include <iostream>
using namespace std;
int findTrailingZeros(int n)
{

int count = 0,sum=0,res=0;

while(n>5)
{
res=n/5;
sum=sum+res;
n=res;
}
return sum;
}
int main()
{
int n,res;
cin>>n;
res=findTrailingZeros(n);
cout<<res;
return 0;

}

#include <iostream>
using namespace std;
int findTrailingZeros(int n)
{
int count = 0,sum=0,r res=n/
sum=sum+res;
n=res;
}
return sum;
}
int main()
{
int n,res;
cin>>n;
res=findTrailingZeros(n);cout
<<res;
return 0;
}

#include <iostream>
using namespace std;
int findTrailingZeros(int n)
{
int count = 0,sum=0,r res=n/
sum=sum+res;
n=res;
}
return sum;
}
int main()
{
int n,res;
cin>>n;
res=findTrailingZeros(n);cout
<<res;
return 0;
}

For calculating the number of zeroes in the given integer we will have to check for n&1 == 0 every time after shifting the bit to the right by 1 and then, if found increment the count and hence, return the count when all the bits have been traversed.

Implementation:

#include<bits/stdc++.h>
using namespace std;
int countno(int r){
    //int r = ~n;
    int count = 0;
    while(r > 0){
        if((r&1) == 0)
            count++;
        r = r>>1;
    }
    return count;
}
int main(){

    int n;
    n = 2;
    cout<<countno(n)<<endl;
    return 0;

}

It can be done as follows.
int num=1034000500;
int count =0;
while(num>0){
int modn=num%10;
num=num/10;
if(modn==0){
count++;
}

}

// count is the result.

for(int i = 0; i < sizeOf(int) * 8 ; i++)
{
x & x-1
count++;
}

Answer = sizeOf(int)*8 - count

Basically number of 1's minus the total possible binary digits.